For each pair $(V,W)$ of vector spaces (over a fixed ground field), let $T(V,W)$ be their tensor product, and $F(V,W)$ the vector space of bilinear forms on $V^*\times W^*$. One checks that
(a) there is a unique linear map $e(V,W)$ from $T(V,W)$ to $F(V,W)$ satisfying
$$
\big(e(V,W)(v\otimes w)\big)(f,g)=f(v)g(w)
$$
for all $v\in V,w\in W,f\in V^*,g\in W^*$,
(b) $e(V,W)$ is injective,
(c) $T$ and $F$ are functors,
(d) $e$ is a natural transformation from $T$ to $F$,
(e) $T,F$ and $e$ are compatible (in an obvious sense) with finite direct sums.
Claim 1: $e(V,W)$ is surjective $\iff$ the cardinal number $\dim(V)\dim(W)$ is finite.
In view of (b), implication "$\Leftarrow$" follows by dimension counting. It suffices thus to prove the non-surjectivity when $V$ is infinite dimensional and $W$ nonzero. Writing $W$ as $W_1\oplus W_2$ with $\dim W_1=1$ and using (b), we are reduced to
Claim 2: if $V$ is infinite dimensional, then the canonical embedding $V\to V^{**}$ is not surjective.
To prove this, we'll use an embedding of $V$ in $V^*$, and an embedding of $V^*$ in $ V^{**}$. None of these two embeddings will be canonical, but their composition will.
Choose a basis $B$ of $V$, and identify $V$ to the space $K^{(B)}$ of finitely supported $K$-valued functions on $B$. Then $V^*$ can be identified to the space $K^B$ of all $K$-valued functions on $B$. Similarly, we can identify $V^*$ to $K^{(B\sqcup C)}$, where $C$ is a set and $\sqcup$ means "disjoint union". As $B$ is infinite, $C$ is nonempty. Using the same trick once more, we can identify $V^{**}$ to $K^{(B\sqcup C\sqcup D)}$, where $D$ is a nonempty set. Then the natural embedding of $K^{(B)}$ in $K^{(B\sqcup C\sqcup D)}$, which is clearly not surjective, corresponds to the natural embedding of $V$ in $V^{**}$. This completes the proof.
The general situation is the following: $A$ is a commutative ring and $V_1, V_2, W_1, W_2$ are $A$-modules
If one of the ordered pairs $(V_1,V_2)$, $\,(V_1,W_1\,$ or $\,(V_2,W_2)$ consists of finitely generated projective $A$-modules, the canonical map is an isomorphism (Bourbaki, Algebra, Ch. 2 ‘Linear Algebra’, §4, n°4, prop. 4).
Over a field, all modules are projective since they're free. So the answer is ‘yes’.
In the present case, here is a sketch of the proof:
Let $K$ be the base field. As $V_1\simeq K^m$ for some $m>0$ and similarly $V_2\simeq K^n$, and the $\operatorname{Hom}$ and $\,\otimes\,$ functors comute with direct sums, we may as well suppose $V_1=V_2=K$.So we have to prove:
$$\widehat\Psi\colon\operatorname{Hom}(K,W_1)\otimes\operatorname{Hom}(K,W_2)\to\operatorname{Hom}(K\otimes K,W_1\otimes W_2)$$
is an isomorphism.
This results basically that for any vector space $V$ we have an isomorphism
\begin{align*}
\operatorname{Hom}_K(K,V)&\simeq V\\
\phi&\mapsto\phi(1)
\end{align*}
In detail, just consider the following commutative diagram:
Best Answer
Two finite dimensional vector spaces $V$ and $W$ are isomorphic if and only if their dimensions are the same, since all vector spaces are free.
Hence, your question amounts to calculating the dimension of $\operatorname{Hom}_k(V,W)$, which is $\dim(V)\dim(W)$ and that of $V\otimes W^*,$ which is $\dim(V)\dim(W^*)=\dim(V)\dim(W)$.